FatMouse's Speed

题目：

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing. 

Input

Input contains data for a bunch of mice, one mouse per line, terminated by end of file. 

The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice. 

Two mice may have the same weight, the same speed, or even the same weight and speed. 

Output

Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that 

W[m[1]] < W[m[2]] < ... < W[m[n]] 

and 

S[m[1]] > S[m[2]] > ... > S[m[n]] 

In order for the answer to be correct, n should be as large as possible. 
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one. 

Sample Input

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900

Sample Output

4
4
5
9
7

 

题意：

题目大意是找到一个最多的老鼠序列，使得序列中的老鼠的体重满足递增，相应老鼠的速度满足递减。

 

思路：

把体重排序，然后按照体重找到最长递减子序列即可。注意此题还需要输出找到的序列中的老鼠的最原始的标号，因此不仅要在刚开始的时候把每个老鼠的最初的序号记下来，还要在进行状态转移的时候把当前的老鼠的位置标记下来。

 

代码：

#include
#include
#include
#include
#include 
using namespace std;
struct Mice{
	int w,v,id;
}mice[1010];
int dp[1010];
int pre[1010];
bool cmp(Mice a,Mice b)
{
	return a.w>b.w;
} 
 
int main()
{
	int s=0;
	while(scanf("%d%d",&mice[s].w,&mice[s].v)!=EOF)
	{
		mice[s].id=s+1;
		s++;
	}
	sort(mice,mice+s,cmp);
	memset(pre,-1,sizeof(pre));
	for(int i=0;imice[i].w&&mice[j].vans)
		{
			ans=dp[i];
			p=i;
		}		
	}
	printf("%d\n",dp[p]);
	while(p!=-1)
	{
		printf("%d\n",mice[p].id);
		p=pre[p];
	}
return 0;
}